Lecture note gas turbine cycle

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GAS TURBINE CYCLES

Introduction

Chapter Objective

• To carry out first law analysis on a gas turbine plant

in which the working fluid is assumed as a perfect

gas.

How a Gas Turbine Works?

• It is a heat engine in which a pressurized hot gas

spins a gas turbine, thus producing mechanical work.

• This pressurized gas produced by burning fuels such

as propane, natural gas, kerosene or jet fuel.

• The heat that comes from burning fuel expands air,

and the high speed rush of this hot air spins the

turbine.•1-Oct-13 •Gas Turbine Cycle •2

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Gas Turbine Combined Cycle

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The Use of Gas Turbine

Gas turbine plants are widely used in the following

engineering fields:

1. Aircraft propulsion system

2. Electric power generation

3. Marine vehicle propulsion

4. Combined-cycle power plant (with steam

power plant)


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Aircraft engine


Front view of the aircraft engine

Aircraft engine sketch

Gas Turbine Animation


Gas Turbine Animation.mp4

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Note:

The processes taking place in an actual gas turbine

plant are complicated. To carry out

thermodynamics study on the system, we will

develop a simplified model of the system.


Types of Gas Turbine Cycles

There are two types of gas turbine cycle: Brayton/Joule Cycle

and Atkinson Cycle.

Brayton Cycle

Heat added and rejected is at constant pressure.

Atkinson Cycle

1. Heat added at constant volume, therefore

a. It needs valve to control gas flow to ascertain constant volume

b. Complicated

2. Heat rejected at constant pressure

3. Not popular

4. Out of the SKMM2423 scope.


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Brayton Cycle

• It is an ideal and closed type cycle.

• It is used as a reference cycle, where actual

cycle can be compared with this cycle.

• The working fluid is merely air.

• The basic components comprise of air

compressor, heat exchanger and gas turbine.

•1-Oct-13 •Gas Turbine Cycle•9

Brayton/Closed Cycle

The cycle comprises of 4 internal

reversible processes.

• Process 1-2: isentropic

compression

• Process 2-3: isobaric heat addition

• Process 3-4: isentropic expansion

• Process 4-1: isobaric heat rejection

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•10

1

3

4

2

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Cycle Analysis - Thermal Efficiency

By using steady flow energy equation (sfee), this closed ideal

cycle can be analyzed as follows,

( )( )( )( )1212

4334

1441

2323

TTcww

TTcww

TTcqq

TTcqq

pcom

ptur

pout

pin

−==

−==

−==

−==



2

4

1

3

i.e.

( )( )23

1411

TTc

TTc

q

q

q

qq

q

w

wwqqw

p

p

in

out

in

outin

in

netth

compturoutinnet

−/

−/−=−=

−==

−=−=

η


( )( )23

141TT

TTth −

−−=η

(1)

First law of thermodynamic states, net heat received by any

cyclic device is the same with the net work produced, i.e.,


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γγ 1

4

3

4

3

−

=p

p

T

T

32 pp = 14 pp =


γγ 1

1

2

1

2

−

=p

p

T

T

But, for isentropic processes,

and

Because

and

Therefore,

prp

p

p

p==

4

3

1

2

(2)

1

3

4

2


2

4

1

3

ratio pressure=pr

( )4.1air ratioheat specific ===v

p

c

cγ

γγ 1

43

−

⋅= prTT


•14

Where,

So,

and

andγγ 1

12

−

⋅= prTT

2

4

1

3


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•15

Therefore,

(3)γγ 1

4

3

1

2−

== prT

T

T

T

2

4

1

3


( ) ( ) γγ 1

1423

−

−=− prTTTT


So,

or

( )( )

γγ 1

23

14 1−=

−−

prTT

TT


2

4

1

3

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( )( )

γγη1

23

14 111 −−=

−−

−=

p

th

rTT

TT


So,

(4)


2

11T

Tth −=η

( )( ) 2

1

2

32

1

41

23

14

1

1

T

T

T

TT

T

TT

TT

TT=

−

−

=−−


•18

Also,

Therefore,

(5)

Note: Only valid for ideal closed cycle.


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( ) ( )( )43

1243

34

1234

TTc

TTcTTc

w

ww

w

ww

p

pp

turbine

netr −

−−−=

−==


Cycle Analysis - Work Ratio, wr,


2

4

1

3

( )( )43

121TT

TTwr −

−−=


Cycle Analysis - Work Ratio, wr,

ie.,


2

4

1

3

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−

−

−=

−

−

−=−

−−

−

−

1

.1

1

11

1

11

3

11

1

13

1

1

γγ

γγ

γγ

γγ

γγ

p

pp

p

p

r

rT

rrT

r

T

rT

w

γγ 1

12 .−

= prTT


•21

We know that,

andγγ 1

34 −=

pr

TT

Therefore,


γγ 1

3

1 .1−

−= pr rT

Tw


•22

or,

(6)


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Modifications of Brayton Gas Turbine Cycle

Brayton/Joule

Cycle

(Closed & Ideal)

Open Cycle

BasicTwo-Stage

Expansion Two-Stage

Compression

with

Intercooler

Two-Stage

Compression

+

Two-Stage

Expansion

+

Reheating

Basic

+

Two-Stage

Expansion

+

Two-Stage

Compression

+

Reheating

+

The Use

Of

Heat Exchanger•1-Oct-13 •Gas Turbine Cycle

•23

Basic Gas Turbine


Basic Components

The basic components of a gas turbine plant working

on an open cycle. The plant comprises of:

1. Air compressor (centrifugal or axial-flow type)

2. Combustion chamber

3. Gas turbine

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Basic Gas Turbine

p2

2’

’

Basic Gas Turbine


Process Description

1-2: Compression process

Atmospheric air at pressure p1 and

temperature T1 is induced and

compressed adiabatically to higher

pressure p2 and temperature T2. The

process is not reversible, thus, not

isentropic.

2-3: Combustion process

Fuel is injected into the air stream. The mixture of air and fuel

is burned at constant pressure inside a combustion chamber

(CC), thus producing hot combustion gases.

p2

2’

’

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Process Description

3-4: Expansion process

The hot combustion gases expands

through the gas turbine. The

process is assumed adiabatic but

not reversible, thus, not isentropic.

Mechanical work is produced by the

turbine. Part of this work is used to

drive the compressor.

The air exiting the turbine is exhausted to atmosphere. New

fresh air is induced into the compressor and the processes are

repeated.

p2

2 ’

’

Basic Gas Turbine


Energy Analysis

• The steady-flow energy equation

(SFEE) is applied to each

component of the gas turbine

plant.

• Neglecting the change in the

kinetic and potential energy of

the working fluid, we have, work

to drive the compressor:

(7)

p2

2’

’

Basic Gas Turbine

( )' 12.C paw c T T= −

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Heat added to the compressed air

(8)

Work produced by the turbine

(9)

Note:

The turbine is connected to the compressor via a common shaft.

Thus, part of the work produced by the turbine is used to drive the

compressor.

p2

2 ’

’

Basic Gas Turbine

( )'3 2.CC pgq c T T= −

( )'3 4.T pgw c T T= −


•30

Properties of Working Fluid

The air and combustion gases are assumed to have the

following properties:

For air: cpa = 1.005 kJ/kg.K, γa = 1.4

For gas: cpg = 1.110 kJ/kg.K, γg = 1.33

Basic Gas Turbine

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Isentropic Efficiency

The compression and expansion processes are not

isentropic. The isentropic efficiency of the compressor and

the turbine is defined as follows,

Compressor:

(10)

Turbine:

(11)

p2

2 ’

’

Basic Gas Turbine

( )( )'

2 1

,

12

is C

T T

T Tη

−=

−

( )( )

'3 4

,

3 4

is T

T T

T Tη

−=

−

γγγ

γ1

1

1

1

212

−

−

=

= prT

p

pTT

γγ

γγ

1

13

1

3

434

−

=

=

−

prT

p

pTT


Isentropic Process

The isentropic process between path 1-2 is

And the isentropic process between path 3-4 is

where the values of γ for air

and gas are different.

Basic Gas Turbine

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Thermal efficiency

Thermal efficiency of the basic gas turbine

cycle is

Cycle Performance

The performance of the basic gas turbine cycle is measured by the

following criteria:

(12)

p2

2 ’

’

Basic Gas Turbine

( ) ( )( )' '

'

3 14 2

3 2

net T Cth

in CC

pg pa

th

pg

w w w

q q

c T T c T T

c T T

η

η

−= =

− − −=

−


Work ratio

The work ratio of the plant is

(13)

p2

2 ’

’

Cycle Performance

The performance of the basic gas turbine cycle is measured by the

following criteria:

Basic Gas Turbine

( )( )

'

'

12

3 4

1

1

net T C Cr

T T T

pa

r

pg

W w w ww

W w w

c T Tw

c T T

−= = = −

−= −

−

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Assumptions

The following assumptions are made on the simplified model

of the gas turbine plant:

1. The mass of fuel injected into the air is ignored since the air

fuel ratio is usually large.

2. The mass flow rate of the working fluid is considered

constant.

fag mmm /+=


Basic Gas Turbine

Example 1

A gas turbine has an overall pressure ratio of 5/1 and a

maximum cycle temperature of 550oC. The turbine drives a

compressor and an electric generator, the mechanical

efficiency of the drive being 97%. The ambient temperature is

20oC and the isentropic efficiencies of the compressor and

turbine are 0.8, and 0.83 respectively. Calculate the power

output in kilowatts for an air flow of 15 kg/s. Calculate also

the thermal efficiency and the work ratio. Neglect changes in

kinetic energy, and the loss of pressure in the combustion

chamber. Please use the following assumptions: for air,

cp = 1.005 kJ/kg and γ = 1.4, for gas, cp = 1.15 kJ/kg and

γ = 1.333.


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Basic Gas Turbine

p2

2’

’

Two-Stage Expansion Process

• It is more practical to expand the hot gases in two

turbine stages.

• The high-pressure (HP) turbine is solely used to

drive the compressor.

• The low-pressure (LP) turbine produces net power

output.


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2

2’

4

4’

5 5’


Energy Analysis

Neglecting changes in the kinetic and potential energy of the

working fluid,

Compressor Work

The high-pressure turbine is dedicatedly used to drive the

compressor, therefore

Work required to drive

the air compressor =

Work produced by the

high-pressure (HP)

turbine


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Energy Analysis

i.e. wC = wT,hp

or

(14)

where cpa and cpg are specifics heats at constant pressure for

the air and hot gases respectively.


( ) ( )2' 1 3 4'. .pa pgc T T c T T− = −


Note:

Given the value of T1, maximum cycle

temperature T3 and the overall

pressure ratio, (p2/p1), eq. (14) can

be used to determine the

temperature T4 at the inlet to the

low-pressure (LP) turbine and the

pressure ratio (p2/p4) across the

high-pressure (HP) turbine.


2

2

’

4

4’

55

’

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Net work output

The low-pressure turbine produces

the net work output for the plant,

thus,

(15)


2

2

’

4

4’

55

’( ),

4' 5'.

net T lp

net pg

w w

w c T T

=

= −

A gas turbine unit takes in air at 17oC and 1.01 bar with an

overall pressure ratio 8:1 and a maximum cycle temperature

of 650oC. The compressor is driven by the high-pressure

(HP) turbine and the low-pressure (LP) turbine drives a

separate power shaft. The isentropic efficiencies of the

compressor, the HP turbine and the LP turbine are 0.8, 0.85

and 0.83 respectively and the mechanical efficiency of both

shafts is 0.95. Calculate the:

a) pressure and temperature at the inlet of LP turbine,

b) net power output for each kg/s mass flow rate,

c) work ratio of the plant, and

d) thermal efficiency of the cycle.


Example 2

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2

2’

4

4’

5 5’

Two-Stage Compression Process


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• In real gas turbine plant, the compression process is carried

out in more than one stage.

• An intercooler is used between each stage.

4

4’

A

A’

6’6

2

2’


Reasons

• The work required to drive the compressor can be reduced.

• Since the work output of the low-pressure (LP) turbine is

unchanged, the work ratio of the plant is increased.

Analysis of the Cycle

The work required to drive compressor with inter-cooling,

(16)


( ) ( ), 2 ' 1 4' 3. .C ic pa paw c T T c T T= − + −∑

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( ) ( )'2'1'2 .. TTcTTcw ApapaC −+−=∑

( ) ( )'2'3'4 .. TTcTTc Apapa −<−

∑ ∑< CicC ww ,

•1-Oct-13

•Gas Turbine Cycle•49

The work required to drive compressor without intercooling,

Since the pressure lines on the T-s

diagram diverge to the right,

Therefore, the compressor work

input with intercooling is less than

the work input without

intercooling, i.e.


44’

AA’

6’6

2

2’

3

4

1

2

p

p

p

p=

13 TT =•1-Oct-13 •Gas Turbine Cycle

•50

Minimum Compressor Work

The work input with intercooling, ΣwC,ic

will be MINIMUM when:

1. The pressure ratio in each stage

is equal, i.e.

2. The intercooling process is complete, i.e. the temperature of the air

entering between the stages the high-pressure (HP) and the low-

pressure (LP) stage is the same, i.e.


44’

A

A’

6’6

2

2’

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T

CTr

w

www

−=


Advantage

Two-stage compression with inter-

cooling between the stages reduces the

work required to drive the compressor.

Recall,

Since wC is reduced while wT remains

unchanged, the work ratio of the plant

increased.


44’

AA’

6’6

2

2’

( )'45, . TTcq pgicCC −=

( )'5. ApgCC TTcq −=


Disadvantage

The heat added to the air during

combustion with 2-stage compression

and inter-cooling is

and that without 2-stage & inter-cooling

is

From T-s diagram, (T5 – T4’) > (T5 – TA’). Therefore the heat added to the air

increases with 2-stage compression with inter-cooling.

Thus, thermal efficiency of the cycle decreases.

44’

AA’

6’6

2

2’


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•27

'4'2

31

TT

TT

≠

≠3

4

1

2

p

p

p

p=

3

4

1

2

p

p

p

p≠


•53

Comparison Between Complete & Incomplete Inter-cooling

T1 = T3

T2’ = T4’

s

T

1

22’

3

4 4’6

5

6’

T

s

1

22’

3

4

4’

6

5

6’


A gas turbine unit has a pressure ratio of 10/1 and a

maximum temperature of 700oC. The compression process is

carried out in 2 stages and inter-cooling is used between the

stages. The isentropic efficiencies of the compressors and the

turbine are 0.82 and 0.85 respectively. The air enters the

compressor at 15oC at the rate 15 kg/s. Calculate,

a) the power output of an electric generator geared to

the turbine;

b) the work ratio;

c) the thermal efficiency.

Assume the conditions for minimum compressor works


Example 3

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Two-Stage Expansion & Reheating


With 2-stage expansion, the gas exiting the high-pressure (HP) turbine can

be reheated in a second combustion chamber, before expanding the low-

pressure (LP) turbine.

22’

4’

4

A A’

6

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( )''4, . ApgLPt TTcw −=


•57

Analysis of the cycle

Neglecting any mechanical losses, the

work produced by the low-pressure

(LP) turbine WITH reheating,

(17)

and the work produces WITHOUT

reheating,

22’

4’

4

A A’

6


( ), , 5 6'.t LP rh pgw c T T= −

( ) ( )''4'65 ATTTT −>−


•58

Analysis of the cycle

Hence, reheating between the turbine

stages increases the work output of the

low-pressure (LP) turbine.

The constant pressure lines diverge to

the right on the T-s diagram, so


22’

4’

4

A A’

6

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•30

∑∑∑ −=

−=

t

c

t

ct

rw

w

w

www 1


Advantage

The work ratio of the plant is given

by

The total turbine work, Σwt increases with reheating between the turbine

stages. Therefore the work ratio of the plant increases.


22’

4’

4

A A’

6


cpg(T5 – T4’) is the additional heat added to the working fluid in the second

combustion chamber.

Clearly, reheating between the turbine stages increases the amount of heat

supplied in the combustion chambers.

This causes the thermal efficiency of the plant to decrease.

Disadvantage

The heat added to the working fluid in

the combustion chambers,

(18)


22

’

4

’

4

A A’

6( ) ( )3 2' 5 4'CC pg pgq c T T c T T= − + −∑

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A gas turbine unit is used in a marine ship consists of a compressor, two

combustion chambers and two stages of turbines. The compressor is

driven by the high-pressure (HP) turbine and the low-pressure (LP)

turbine drives the ship rotor. The overall pressure ratio is 10:1. The

atmosphere air enters the compressor at 1.01 bar and 30oC with the rate

of 20 kg/s. The gas enters the HP turbine at 900oC and enters the LP

turbine at 700oC. The mechanical efficiency for both shafts is 90%. The

isentropic efficiency for both turbines is 85%. The power received by the

rotor is 2590 kW.

a) Show the plant components arrangement and all the processes

on a T-s diagram,

b) Calculate the intermediate pressure between the HP turbine and

LP turbine,

c) Calculate the compressor power,

d) Calculate the isentropic efficiency of the compressor,

e) Thermal efficiency of the cycle.


Example 4



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•32

A gas turbine power plant consists of 2-stages compressor and 2-stages turbine. The high-pressure and low-pressure compressorare driven by the high-pressure (HP) turbine and the low-pressure(LP) turbine drives a separate power shaft. The low-pressurecompressor takes in air at 100 kPa and 27oC with the mass flowrate of 5.80 kg/s. The compressors have equal pressure ratios andinter-cooling is complete between stages. The temperature of thegases at entry to the HP turbine is 1327oC and the gases arereheated to 1127oC at 300 kPa after expansion in the first turbine.The overall pressure ratio is 10. The isentropic efficiency of eachcompressor stage and each turbine stage is 80 %. Calculate;

a) net power output (kW),

b) work ratio of the plant,

c) thermal efficiency of the cycle.

Sketch the schematic diagram of the plant and all the processes ona T-s diagram.


Example 5


HPC

CC

LPT

Exhaust

Net power output

8’

4’3

HP

T

5CC

6’ 7

4

4

’

2

‘

‘ ‘

Two Stage Compression & Two Stage

Expansion

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•33

The Use of a Heat Exchanger




• The gases exiting the low-pressure (LP) turbine is still has a high

temperature.

• The heat energy contained in the exhaust gases can be utilized for

improving the thermal efficiency of the cycle.

• One scheme is by using a heat exchanger unit to preheat the air leaving

the compressor, before it enters the combustion chamber.

Net power

output

’

’Ideal heat

exchanger

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•34

6'2 TT = '53 TT =


Ideal vs Actual Temperatures

In an ideal cycle, temperatures and

In an actual cycle, temperatures '53 TT <6'2 TT < and

The finite temperature difference is

required, between the compressed air

and the exhaust gases, for the heat

transfer process to take place.


’

’


Advantage of Heat Exchanger

When the compressed air is preheated before entering the combustion

chamber, the amount of heat qCC required to raise the temperature of the

working fluid from T2’ to T4, is reduced.

Heat only required to raise the temperature of

the air from T3 to T4. This leads to saving in fuel

consumption and hence the operating cost of

the plant.

Assuming that net work output of the plant

remains unchanged, the thermal efficiency of

the cycle increases by using the heat exchanger.

Actual heat exchanger

’

’


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•35


Heat Energy Balance

Assuming the heat exchanger is well insulated (no heat loss), we have

Heat transferred

FROM the exhaust

gases

=

Heat transferred

TO the compressed

air

i.e. (19)


( ) ( )5' 6 3 2'g pg a pam c T T m c T T− = −

difference re temperatuavailable Maximum

air theof rise eTemperatur=hetr


Heat Exchanger Performance

The performance of the heat exchanger is measured by it’s

thermal ratio, defined as,

(20)

i.e.


( )( )

3 2 '

5 ' 2 '

he

T Ttr

T T

−=

−

•1/10/2013

•36


•71

Heat Exchanger Performance

Without heat exchanger :

With heat exchanger :

(21)

(22)

Actual heat exchanger

’

’


( )4 2'in pgq c T T= −

( )4 3in pgq c T T= −


Conditions for Use

To use the heat exchanger, there must

be a sufficiently large temperature

difference between the exhaust gases

and the compressed air.

The use of heat exchanger will not be

feasible when the temperature of the

exhaust gases is lower than the

temperature of the compressed air

leaving the compressor.


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•37


You are required to carry out an energy audit on a gas turbine plant. In

this plant the high pressure turbine (HPT) drives a compressor and the

low pressure turbine (LPT) drives an electric generator (EG).

Atmospheric air, after being compressed, enters a first combustion

chamber (CC1) and the hot gas is then expanded in the HPT. The gas is

reheated in a second combustion chamber (CC2) before it is expanded in

the LPT. Based on the true measurements made, followings are the data

recorded by your Technical Assistant:

o output power of generator, 1080 kW;

o mass flow-rate of air, 6.85 kg/s;

o compressor pressure ratio, 6.3 and HPT pressure ratio, 2.9;

o compressor inlet and outlet air temperatures, 20oC and 252oC;

o HPT inlet and outlet gas temperatures, 701oC and 487oC;

o LPT inlet and outlet gas temperatures, 631oC and 486oC.

Example 6


Sketch all processes on a T-s diagram and determine the following values,

for you to prepare the audit report:

i. compressor isentropic efficiency;

ii. HPT isentropic efficiency;

iii. HPT/compressor mechanical efficiency;

iv. LPT isentropic efficiency;

v. LPT/EG mechanical efficiency;

vi. plant thermal efficiency, and

vii. in the same report, you are proposing that a heat exchanger with a

thermal ratio of 0.78, be installed to preheat the compressed air by

the exhaust gas. To support your proposal what are:

(a) the percentage saving on heat supply by CC1, and

(b) the new value of the plant thermal efficiency.

Please use the following assumptions: for air, cp = 1.005 kJ/kg and γ = 1.4

for gas, cp = 1.15 kJ/kg and γ = 1.333

Example 6

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•38

A 5000 kW gas turbine generating set operates with two

compressor stages with inter-cooling between stages; the

overall pressure ratio is 9/1. A HP turbine is used to drive the

compressors, and a LP turbine drives the generator. The

temperature of the gases at entry to the HP turbine is 650°C

and the gases are reheated to 650°C after expansion in the

first turbine. The exhaust gases leaving the LP turbine are

passed through a heat exchanger to heat the air leaving the HP

stage compressor. The compressors have equal pressure ratios

and inter-cooling is complete between stages.


Example 7

The air inlet temperature to the unit is 15°C. The isentropic

efficiency of each compressor stage is 0.8 and the isentropic

efficiency of each turbine stage is 0.85; the heat exchanger

thermal ratio is 0.75. A mechanical efficiency of 98% can be

assumed for both the power shaft and the compressor turbine

shaft. Neglecting all pressure losses and changes in kinetic

energy, calculate: (i) the cycle efficiency; (ii) the work ratio;

(iii) the mass flow rate. For air take cp = 1.005 kJ/kg.K and γ =

1.4, and for the gases in the combustion chamber and in the

turbines and heat exchanger take cp = 1.15 kJ/kg.K and γ = 1.33.

Neglect the mass of fuel.


Example 7

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In a gas turbine generating station the overall pressure ratio is

15/1. The compression is performed in three stages with pressure

ratios of 3/1, 2.5/1 and 2/1 respectively. The air inlet temperature

of the plant is 30oC and inter-cooling between stages reduces the

temperature to 45oC. The high pressure (HP) turbine drives all the

compressors and the low pressure (LP) turbine drives the

generator. The gases leaving the LP turbine are passed through a

heat exchanger which heats the air leaving the HP compressor to

250oC. The temperature at inlet to the HP turbine is 700oC, and

reheating between turbine stages raises the temperature to

650oC. The isentropic efficiency of each compressor stage is 0.85,

and the isentropic efficiency of each turbine is 0.88. Take the

mechanical efficiency of each shaft as 98%. The air mass flow rate

is 120 kg/s.


Example 8

Neglecting pressure losses and changes in kinetic energy, and

taking the specific heat of water as 4.19 kJ/kgK, calculate,

i. The power output (kW)

ii. The cycle efficiency

iii. The mass flow rate of cooling water for the intercoolers

if its temperature rise must not exceed 20 K

iv. The heat exchanger thermal ratio.

Assume that cp and γ may be taken as 1.005 kJ/kgK and 1.4 for

air, and as 1.15 kJ/kgK and 1.33 for combustion and expansion

processes.


Example 8

Education

Lecture note gas turbine cycle